1,1,2,2- tetrachloropropane was heated with zinc dust and the product was bubbled through ammonical `AgNO_3`. What is the weight of precipitate obtained?

Options

A. 30.0g
B. 29.4g
C. 28.0g
D. 25.7g

Explanation

To solve the problem step by step, we will follow the reactions and calculations involved in the process.

### Step 1: Identify the starting compound
The starting compound is 1,1,2,2-tetrachloropropane, which has the molecular formula C3H4Cl4.

### Step 2: Determine the molecular weight of 1,1,2,2-tetrachloropropane
To calculate the molecular weight:
- Carbon (C): 12 g/mol (3 Carbon atoms = 3 × 12 = 36 g/mol)
- Hydrogen (H): 1 g/mol (4 Hydrogen atoms = 4 × 1 = 4 g/mol)
- Chlorine (Cl): 35.5 g/mol (4 Chlorine atoms = 4 × 35.5 = 142 g/mol)

Total molecular weight = 36 + 4 + 142 = 182 g/mol

### Step 3: Reaction with zinc dust
When 1,1,2,2-tetrachloropropane is heated with zinc dust, it undergoes a dehalogenation reaction to form propine (C3H4), which has a molecular formula of C3H4.

### Step 4: Determine the molecular weight of propine
To calculate the molecular weight of propine:
- Carbon (C): 12 g/mol (3 Carbon atoms = 3 × 12 = 36 g/mol)
- Hydrogen (H): 1 g/mol (4 Hydrogen atoms = 4 × 1 = 4 g/mol)

Total molecular weight of propine = 36 + 4 = 40 g/mol

### Step 5: Reaction with ammonical AgNO3
The propine is then treated with ammonical silver nitrate (AgNO3). This reaction will lead to the formation of a silver precipitate (AgC≡C−R) where R is the remaining part of the propine molecule.

### Step 6: Calculate the weight of the precipitate formed
From the stoichiometry of the reaction:
- 182 g of 1,1,2,2-tetrachloropropane produces 147 g of precipitate.

Now, we can set up a proportion to find out how much precipitate is formed from 36.4 g of 1,1,2,2-tetrachloropropane:
\[ \text{Weight of precipitate} = \left( \frac{147 \text{ g}}{182 \text{ g}} \right) \times 36.4 \text{ g} \]

Calculating this gives:
\[ \text{Weight of precipitate} = \left( \frac{147}{182} \right) \times 36.4 \approx 29.4 \text{ g} \]

### Conclusion
The weight of the precipitate obtained is approximately **29.4 grams**.

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